You can actually prove a stronger result easily using vectors. The diagonals of a parallelogram bisect one another. The rhombus is just a special case of a parallelogram with all sides being equal.

Let the parallelogram be drawn on a Cartesian plane as shown in the diagram, and the sides labelled as vectors as shown. You can see that one diagonal is [tex]\vec a + \vec b[/tex] and the other is [tex]\vec b - \vec a[/tex]

Let [tex]\vec{WO}[/tex] be [tex]k_1(\vec a + \vec b)[/tex]

and

[tex]\vec{OZ}[/tex] be [tex]k_2(\vec b - \vec a)[/tex]

where [tex]k_1, k_2[/tex] are some scalars (which we are to determine).